A small bar magnet experiences a torque of 0.016 Nm when placed with its axis at 30° with an external field of 0.04 T. If the bar magnet is replaced by a solenoid of cross-sectional area of `1 cm^2` and 1000 turns but having the same magnetic moment as that of bar magnet, then the current flowing through the solenoid is

Options

A. `0.083`
B. `0.16`
C. `0.25`
D. `0.3`

Answer

A

Explanation

Given,
For a small bar magnet,
Torque, `tau = 0.016NM`
`there = 30^(@)`
External field, ` = 0.04T`
Area of solenoid, `A = 1 cm ^(2) =10 ^(-4)m ^(2)`
Number of turns of solenoid, `N = 1000`
Magnetic moment of solenoid is given by,
`m = NIA`
`implies I = (m)/(NA)" "...(1)`
But,
`m = (tau)/(B sin theta ) (because tau = mB sin theta)`
`implies I = (tau)/((B sin theta )/(NA))`
`= (tau)/( NA B sin theta)`
` = ( 0.016)/(1000 xx 10 ^(-4) xx 0.04 xx sin 30)`
`= (0.016)/(0.004 xx(1)/(2)) = (0.016)/(0.002) = (16)/(2)`
`therefore I = 8 A`